How many moles of sodium bicarbonate is needed to neutralize 0.8 ml of sulphuric acid?

n NaHCO3 = 9.6 E-3 mol

Explanation:

balanced reaction:

2 NaHCO3 (s) + H2SO4 (ac) ↔ Na2SO4 (ac) + 2 CO2 (g) + 2 H2O (l) assuming a concentration of H2SO4 6M ... normally worked in the lab

⇒ n H2SO4 = 8 E-4 L * 6 mol/L = 4.8 E-3 mol H2SO4

according to balanced reaction, we have that for every mol of H2SO4 there are two mol of NaHCO3 (sodium bicarbonate)

⇒ mol NaHCO3 = 4.8 E-3 mol H2SO4 * (2 mol NaHCO3 / mol H2SO4)

⇒, mol NaHCO3 = 9.6 E-3 mol

So 9.6 E-3 mol NaHCO3, are the minimun moles necessary to neutralize the acid.